A ball is thrown from the top of a tower in vertically upward direction. Velocity at a point h m below the point of projection is twice of the velocity at a point h m above the point of projection. Find the maximum height reached by the ball above the top of tower.

Options

A. 2 h
B. 3 h
C. `(5//3) h`
D. `(4//3) h`

Explanation

To solve the problem, we will use the equations of motion under uniform acceleration due to gravity. Let's denote:

- \( u \) = initial velocity of the ball (at the top of the tower)
- \( v_1 \) = velocity of the ball at a height \( h \) below the point of projection
- \( v_2 \) = velocity of the ball at a height \( h \) above the point of projection
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))

### Step-by-Step Solution:

1. **Identify the velocities at the specified points:**
- The velocity at a point \( h \) below the projection point (height = 0) can be expressed using the first equation of motion:
\[ v_1^2 = u^2 - 2g(-h) \quad \text{(since the ball is moving upwards)} \]
Simplifying this gives:
\[ v_1^2 = u^2 + 2gh \]

- The velocity at a point \( h \) above the projection point (height = 0) can be expressed similarly:
\[ v_2^2 = u^2 - 2gh \]

2. **Use the given condition:**
According to the problem, the velocity at the point \( h \) below is twice that at the point \( h \) above:
\[ v_1 = 2v_2 \]
Squaring both sides gives:
\[ v_1^2 = 4v_2^2 \]

3. **Substituting the expressions for \( v_1^2 \) and \( v_2^2 \):**
From the equations derived in step 1, we have:
\[ u^2 + 2gh = 4(u^2 - 2gh) \]

4. **Expanding and simplifying:**
Expanding the right side:
\[ u^2 + 2gh = 4u^2 - 8gh \]
Rearranging gives:
\[ 2gh + 8gh = 4u^2 - u^2 \]
\[ 10gh = 3u^2 \]
Thus, we can express \( u^2 \) in terms of \( g \) and \( h \):
\[ u^2 = \frac{10gh}{3} \]

5. **Finding the maximum height:**
The maximum height \( H \) reached by the ball above the point of projection can be calculated using the formula:
\[ H = \frac{u^2}{2g} \]
Substituting \( u^2 \):
\[ H = \frac{10gh/3}{2g} = \frac{10h}{6} = \frac{5h}{3} \]

6. **Final answer:**
The maximum height reached by the ball above the top of the tower is:
\[ \frac{5h}{3} \]